Parabola

Let $P \equiv(\alpha, 4 \alpha)$

Equation of tangent to $y^2=4 x$

$ \begin{array}{ll} & y=m x+\frac{1}{m} \\ & 4 \alpha=m \alpha+\frac{1}{m} \\ & m^2 \alpha-4 \alpha m+1=0 \\ & m_1+m_2=4 \\ & m_1 m_2=\frac{1}{\alpha} \\ \Rightarrow \quad & \tan \frac{\pi}{3}=\frac{\sqrt{\left(m_1+m_2\right)^2-4 m_1 m_2}}{\left|1+\frac{1}{\alpha}\right|} \\ \Rightarrow \quad & \sqrt{3}=\frac{\sqrt{16-\frac{4}{\alpha}}}{\left|1+\frac{1}{\alpha}\right|} \Rightarrow 3\left(1+\frac{1}{\alpha}\right)^2=16-\frac{4}{\alpha} \\ \Rightarrow \quad & \frac{3(\alpha+1)^2}{\alpha^2}=\frac{16 \alpha-4}{\alpha} \\ \Rightarrow \quad & 3(\alpha+1)^2=\alpha(16 \alpha-4) \\ \Rightarrow \quad & 3 \alpha^2+6 \alpha+3=16 \alpha^2-4 \alpha \\ \Rightarrow \quad & 13 \alpha^2-10 \alpha-3=0 \\ \Rightarrow \quad & \alpha_1+\alpha_2=\frac{10}{13} \end{array} $

⇒ Sum of abscissa of all such point is $\frac{10}{3}$.

Equation of normal

at point (am²,-2am)

is $y=m x-2 a m-a m^3$

Here $a=1$

Let point $p$ be ( $\alpha, \beta$ )

$ \begin{aligned} & \Rightarrow \quad \beta=m \alpha-2 m-m^3 . \\ & \Rightarrow \quad m^3+m(2-\alpha)+\beta=0 \\ & \Rightarrow \quad m_1+m_2+m_3=0 \\ & \sum m_1 m_2=2-\alpha \end{aligned} $

Cyclic

$ \begin{aligned} & m_1 m_2 m_3=-\beta \\ & A \equiv\left(m_1^2,-2 m_1\right) \\ & B \equiv\left(m_2^2,-2 m_2\right) \\ & C \equiv\left(m_3^2,-2 m_3\right) \\ & G \equiv(2,0) \\ & 2=\frac{m_1^2+m_2^2+m_3^2}{3} \\ & 6=\left(m_1+m_2+m_3\right)^2-2 \sum m_1 m_2 \\ & 6=0-2(2-\alpha) \\ & 6=-4+2 \alpha \\ & \alpha=5 \end{aligned} $

Let $P=\left(a t_1^2, 2 a t_1\right) \quad Q=\left(a t_2^2, 2 a t_2\right)$

Normal at $P$ intersect the parabola at $Q$

$ \Rightarrow \quad t_2=-t_1-\frac{2}{t_1} $

Slope of $O P=\frac{2}{t_1}$

Slope of $O Q=\frac{\mathbf{2}}{t_2}$

$ \begin{array}{ll} \Rightarrow & \frac{2}{t_1} \times \frac{2}{t_2}=-1 \Rightarrow t_1 t_2=-4 \\ \Rightarrow & t_1 t_2=-t_1^2-2 \\ \Rightarrow & t_1= \pm \sqrt{2} \\ \Rightarrow & t_1=\sqrt{2} \\ \Rightarrow & t_2=-\sqrt{2}-\frac{2}{\sqrt{2}}=-2 \sqrt{2} \\ \Rightarrow & Q \equiv\left(a t_2^2, 2 a t_2\right) \equiv(8 a,-4 \sqrt{2} a) \end{array} $

$ \begin{aligned} &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \equiv\left(8 \times \frac{5}{4}-4 \sqrt{2} \times \frac{5}{4}\right) \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \equiv(10,-5 \sqrt{2}) \end{aligned} $

$(y-2)^2=3(x-1)$

Let $y-2=Y$ and $x-1=X$

$ \Rightarrow \quad Y^2=3 X . $

and coordinates of end-point of $L, R$

$ X=a $

and $\quad Y= \pm 2 a$

$ \begin{aligned} \Rightarrow & x-1=\frac{3}{4} \text { and } y-2= \pm \frac{3}{2} \\ & x=\frac{7}{4} \text { and } y=2 \pm \frac{3}{2} \\ \Rightarrow & p=\frac{7}{4}, \quad q=\frac{7}{2} \end{aligned} $

Equation of tangent to $(y-2)^2=3(x-1)$ at $\left(\frac{7}{4}, \frac{7}{2}\right)$

$ \begin{aligned} &\begin{aligned} & y-\frac{7}{2}=m\left(x-\frac{7}{4}\right) \\ & m=\left(\frac{d y}{d x}\right)_{\frac{7}{4}, \frac{7}{2}} \\ & 2\left(y-2 \frac{d y}{d x}=3\right. \\ & \frac{d y}{d x}=\frac{3}{2\left(\frac{7}{2}-2\right)}=\frac{3 \times 2}{2 \times 3}=1 \end{aligned}\\ &\text { Equation of tangent, } y-\frac{7}{2}=x-\frac{7}{4}\\ &\begin{array}{ll} & 4 y-14=4 x-7 \\ \Rightarrow & 4 x-4 y+7=0 \end{array} \end{aligned} $

$ \begin{aligned} & y^2=7 x \\ & 4 a=7 \Rightarrow a=\frac{7}{4} \end{aligned} $

Point $(2,0)$

Equation of normal at $\left(a t^2, 2 a t\right)$

$ \begin{aligned} \Rightarrow & & y+t x & =2 a t+a t^3 \\ \Rightarrow & & 0+2 t & =2 a t+a t^3 \\ \Rightarrow & & 2 t & =\frac{7}{2} t+\frac{7}{4} t^3 \end{aligned} $

$ \begin{aligned} \Rightarrow \quad t & =0 \\ t^2 & =\frac{4}{7}\left(2-\frac{7}{2}\right)=\frac{-6}{7} \end{aligned} $

Not possible.

∴ Only one normal.

$y^2=11 x, 4 a=11$

∴ Equation of tangent

$ \begin{aligned} y & =m x+\frac{a}{m} \\ \Rightarrow \quad y & =m x+\frac{11}{4 m} \end{aligned} $

At (1, 4)

$ \begin{aligned} & 4=m+\frac{11}{4 m} \\ \Rightarrow \quad & 16 m=4 m^2+11 \\ \Rightarrow \quad & 4 m^2-16 m+11=0 \\ \Rightarrow \quad & m_1+m_2=4 \\ \Rightarrow \quad & m_1 m_2=\frac{11}{4} \\ & \left(m_1+m_2\right)^2=m_1^2+m_2^2+2 m_1 m_2 \\ \Rightarrow \quad & 16-\frac{11}{2}=m_1^2+m_2^2 \\ \Rightarrow \quad & 21=2\left(m_1^2+m_2^2\right) \end{aligned} $

Given, $y^2=3 x$

Differential w.r.t. $x$, we get

$ 2 y \frac{d y}{d x}=3 \Rightarrow \frac{d y}{d x}=\frac{3}{2 y} $

Slope of tangent at $\left(x_1, y_1\right)=m_t=\frac{3}{2 y_1}$

Slope of normal $=m_n=\frac{-2 y_1}{3}$

∴ Equation of normal

$ \begin{aligned} & y-y_1=\frac{-2 y_1}{3}\left(x-x_1\right) \\ & \text { Put }\left(x_1, y_1\right)=\left(\frac{3}{4}, \frac{3}{2}\right) \\ & \Rightarrow \quad y-\frac{3}{2}=-\frac{2\left(\frac{3}{2}\right)}{3} \cdot\left(x-\frac{3}{4}\right) \\ & \Rightarrow \quad y+x=\frac{3}{4}+\frac{3}{2}=\frac{9}{4} \end{aligned} $

$ \Rightarrow \quad 4 x+4 y=9 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

From $Q(3,3)$

$ \begin{aligned} y-3 & =\frac{-2 \times 3}{3}(x-3) \\ \Rightarrow \quad y+2 x & =9 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

From Eqs. (i) and (ii), we get

$ \begin{array}{ll} & x=\frac{27}{4}, y=\frac{-9}{2} \\ \therefore & R \equiv\left(\frac{27}{4}, \frac{-9}{2}\right) \end{array} $

$y^2=9 x$

$ \Rightarrow \quad a=\frac{9}{4} $

Equation of tangent to parabola

$ \begin{aligned} y & =m x+\frac{a}{m} \\ \Rightarrow \quad y & =m x+\frac{9}{4 m} \end{aligned} $

Since, tangents are drawn from point (1, 5)

$ \begin{array}{lc} \Rightarrow & \quad 5=m+\frac{9}{4 m} \\ \Rightarrow & 4 m^2-20 m+9=0 \\ \Rightarrow & m=\frac{20 \pm \sqrt{400-144}}{8}=\frac{20 \pm 16}{2} \\ \Rightarrow & m=\frac{9}{2}, \frac{1}{2} \end{array} $

$ \begin{aligned} & \therefore \tan \theta=\left|\frac{m_1-m_2}{1+m_1 m_2}\right|=\left|\frac{\frac{9}{2}-\frac{1}{2}}{1+\frac{9}{4}}\right| \\ & \quad=\frac{16}{13} \approx 1.23 \\ & \Rightarrow \quad 1<\frac{16}{13}<\sqrt{3} \\ & \Rightarrow \quad \tan \frac{\pi}{4}<\tan \theta<\tan \frac{\pi}{3} \\ & \therefore \quad \frac{\pi}{4}<\theta<\frac{\pi}{3} . \end{aligned} $

Given, equation of parabola

$ \begin{aligned} & y=x^2-3 x+2 \\ & \Rightarrow \quad y=x^2-3 x+\frac{9}{4}-\frac{9}{4}+2 \\ & \Rightarrow \quad y=\left(x-\frac{3}{2}\right)^2-\frac{1}{4} \\ & \Rightarrow \quad y+\frac{1}{4}=\left(x-\frac{3}{2}\right)^2 \end{aligned} $

which is in the form of

$ \begin{aligned} & (x-h)^2=4 a(y-k) \\ & \therefore \operatorname{Vertex}(h, k)=\left(\frac{3}{2}, \frac{-1}{4}\right) \\ & \therefore \quad Z=\left(\frac{3}{2}, \frac{-1}{2}\right) \end{aligned} $

And $4 a=1 \Rightarrow a=\frac{1}{4}$

$ \begin{aligned} \therefore \text { Focus } S & =(h, k+a) \\ & =\left(\frac{3}{2}, \frac{-1}{4}+\frac{1}{4}\right)=\left(\frac{3}{2}, 0\right) \end{aligned} $

Tangent parallel to the $X$-axis

$\therefore$ Its slope is 0

Now, $y=x^2-3 x+2$

$ \begin{array}{ll} \Rightarrow & \frac{d y}{d x}=2 x-3=0 \\ \Rightarrow & 2 x=3 \Rightarrow x=\frac{3}{2} \end{array} $

$ \begin{array}{rlrl} \therefore & y =\left(\frac{3}{2}\right)^2-3\left(\frac{3}{2}\right)+2 \\ & =\frac{9}{4}-\frac{9}{2}+2 \\ \Rightarrow & y =\frac{9-18+8}{4}=\frac{-1}{4} \\ \therefore & Q =\left(\frac{3}{2},-\frac{1}{4}\right) . \end{array} $

$ \begin{aligned} &\text { The end points of the latus rectum are }\\ &\begin{aligned} (h \pm 2 a, k+a) & \\ & =\left(\frac{3}{2} \pm \frac{1}{2}, 0\right) \\ & =\left(\frac{3}{2}+\frac{1}{2}, 0\right) \text { and }\left(\frac{3}{2}-\frac{1}{2}, 0\right) \\ & =(2,0) \text { and }(1,0) \\ \therefore \quad P & =(2,0) \text { or }(1,0) . \end{aligned} \end{aligned} $

Given, equations of parabola

$ \begin{aligned} & & y^2 & =12 x \\ \therefore & & 4 a & =12 \Rightarrow a=3 \end{aligned} $

$\therefore$ Parametric equation of parabola

$ x=3 t^3, y=2(3) t=6 t $

Any point on the parabola is $P\left(3 t^2, 6 t\right)$. and focus of parabola is $F(3,0)$ Let $Q(x, y)$ be the point that divides the line segment $F P$ in the ratio $m: n$

$ \begin{array}{ll} \therefore & x=\frac{m\left(3 t^2\right)+n(3)}{m+n}=\frac{3 m t^2+3 n}{m+n} \\ \text { And } & y=\frac{m(6 t)+n(0)}{m+n}=\frac{6 m t}{m+n} \\ \Rightarrow & y(m+n)=6 m t \\ \Rightarrow & t=\frac{y(m+n)}{6 m} \\ \therefore & x=\frac{3 m\left(\frac{y(m+n)}{6 m}\right)^2+3 n}{m+n} \end{array} $

$ \begin{aligned} &\begin{aligned} & \Rightarrow \quad x=\frac{\frac{y^2(m+n)^2}{12 m}+3 n}{m+n} \\ & \Rightarrow \quad x=\frac{36 m n+y^2(m+n)^2}{12 m(m+n)} \\ & \Rightarrow \quad 12 m(m+n) x=36 m n+y^2(m+n)^2 \\ & \Rightarrow \quad y^2(m+n)^2=12 m(m+n) x-36 m n \\ & \Rightarrow \quad y^2=\frac{12 m(m+n)}{(m+n)} x-\frac{36 m n}{(m+n)^2} \\ & \Rightarrow \quad y^2=\frac{12 m}{m+n} x-\frac{36 m n}{(m+n)^2} \end{aligned}\\ &\text { Which is in the form of } Y^2=4 A X+B\\ &\therefore \text { Length of latusrectum }=\frac{12 m}{m+n} \end{aligned} $

Given parabola is $y^2=32 x$,

So, $4 a=32$

$ \Rightarrow \quad a=8 $

The point $P$ is $(8,16)$. This in parametric form as $\left(a t^2, 2 a t\right)$

So, $8 t^2=8$

$ \Rightarrow \quad t^2=1 \Rightarrow t= \pm 1 $

And, $2 a t=2(8) t=16 t=16$

$ \Rightarrow \quad t=1 $

So, the perimeter for $P$ is $t_1=1$.

The equation of the normal to the parabola at a point $P\left(a t_1^2, 2 a t_1\right)$ is

$ \begin{aligned} & y+t_1 x=2 a t_1+a t_1^3 \\ & \Rightarrow \quad y+(1) x=2(8)(1)+8(1)^3 \\ & \Rightarrow \quad y+x=16+8=24 \\ & \Rightarrow \quad y=24-x \end{aligned} $

Substitute this into parabola equation

$ \begin{aligned} & y^2=32 x \\ & \Rightarrow \quad(24-x)^2=32 x \\ & \Rightarrow \quad 576-48 x+x^2=32 x \\ & \Rightarrow \quad x^2-80 x+576=0 \end{aligned} $

Since, $x=8$ is one root corresponding to point $P$. Let the other root be $x_Q$.

Using the sum of roots, $8+x_Q=80$

$ \begin{array}{lrl} \Rightarrow x_Q =72 \\ \text { So, } y =24-x_Q \\ =24-72=-48 \end{array} $

So, the coordinates of $Q$ are $(72,-48)$

Now, the equation of the tangent to parabola $y^2=4 a x$ at a point $\left(x_1, y_1\right)$ is

$ \begin{aligned} & y y_1=2 a\left(x+x_1\right) \\ & \Rightarrow y(-48)=2(8)(x+72)=16(x+72 \\ & \Rightarrow-3 y=x+72 \Rightarrow x+3 y+72=0 \end{aligned} $

So, the equation of the tangent drawn at $Q$ to the parabola is $x+3 y+72=0$

Given equation is

$ \begin{array}{ll} & x^2-2 x-4 y+5=0 \\ \Rightarrow & (x-1)^2-4 y+4=0 \\ \Rightarrow & (x-1)^2=4(y-1) \end{array} $

So, $(h, k)=(1,1)$ are the vertex

And $4 a=4$

$ \Rightarrow a=1 $

Since, the parabola open upwards (because the $y$ term is positive and $x$ term is squared), the focus is $(h, k+a)=(1,1+1)=(1,2)$ and the directrix is the line $y=k-a=1-1=0$, which is the $X$-axis.

Since, focal distance of a point on a parabola is its distance from the directrix.

The point $(5,5)$ and directrix is $y=0$.

The distance from a point $\left(x_0, y_0\right)$ to a horizontal line $y=c$ is $\left|y_0-c\right|$.

$\therefore$ Focal distance $=|5-0|=5$

So, the focal distance of the point $(5,5)$ on the parabola $x^2-2 x-4 y+5=0$ is 5 .

Vertex is the mid-point of focus and point of intersection directrix and axis

$ \begin{array}{l} \quad \quad\left(\frac{a+c}{2}, \frac{d+b}{2}\right)=(1, l) \\\\ \Rightarrow \quad a+c=2 \\\\ \text { and } \quad b+d=2 \\\\ \therefore a+b+c+d=4 \end{array} $

Equation of parabola having axis purlild to $ Y $-axis is

$ (x-h)^{2}=4 a(y-k) $

it passes through

$ (1,0):(1-h)^{2}=4 a(-k) $

$ (0,2): h^{2}=4 a(2-k) $

$ (-1,-1):(-1-h)^{2}=4 a(-1-k) $

On subtracting in Eqs (ii) and (iv), we get,

$1+h^{2}-2 h-1-h^{2}-2 h = -4 a K+4 a+4 a K-4 h=4 a $

$ \Rightarrow \quad h=-a $

From Eq. (iii), we get,

$ \begin{array}{rlrl} & a^{2} =4 a(2-k) \\\\ \therefore & a=4(2-k) \\\\ \Rightarrow & h=-8+4 k \end{array} $

On subtracting in Eqs (i) and (iii), we get

$ \begin{aligned} 1+h^{2}-2 h-h^{2} & =-4 a k-8 a+4 a k \\\\ 1-2 h & =-8 a \\\\ 1+2 a & =-8 a \\\\ -10 a & =1 \\\\ a & =-\frac{1}{10} \\\\ h & =\frac{1}{10} \\\\ 4 k & =h+8=\frac{1}{10}+8=\frac{81}{10} \\\\ k & =\frac{81}{40} \end{aligned} $

Put the value of $ B, h, k $ and a in above equation (i), we get

$ \begin{array}{l} \left(x-\frac{1}{10}\right)^{2}=4\left(-\frac{1}{10}\right)\left(y-\frac{81}{40}\right) \\\\ \frac{(10 x-1)^{2}}{10}=-\frac{4}{10}\left(\frac{40 y-81}{40}\right) \\\\ 100 x^{2}+1-20 x=-40 y+81 \\\\ 100 x^{2}-20 x+40 y-80=0 \\\\ 5 x^{2}-x+2 y-4=0 \\\\ a=5 \quad b=-1 c=2 \text { and } d=-4 \\\\ \frac{a d}{b c}=\frac{-20}{-2}=10 \end{array} $

Coordinate of point $ P \equiv\left(\frac{-a t^{2}}{4}, \frac{a t}{2}\right) $

$ \Rightarrow A \equiv\left(\frac{-a t^{2}}{4}, 0\right) \text { and } B \equiv\left(0, \frac{a t}{2}\right) $

Equation of $ A B \Rightarrow \frac{4 x}{-a t^{2}}+\frac{2 y}{a t}=1 $

Point $ Q\left(t_{1}\right)=\left(a t_{1}^{2}, 2 a t_{1}\right) $

Also, point $ Q $ lies on $ A B $

$ \begin{array}{l} \frac{4\left(a t_{1}^{2}\right)}{-a t^{2}}+\frac{2\left(2 a t_{1}\right)}{a t}=1 \\\\ \Rightarrow \quad-4 t_{1}^{2}+4 t_{1} t=t^{2} \\\\ \Rightarrow 4 t_{1}^{2}-4 t_{1} t+t^{2}=0 \\\\ \Rightarrow \quad\left(2 t_{1}-t\right)^{2}=0 \Rightarrow t_{1}=\frac{t}{2} \end{array} $

Focus of given parabola $ (0,3) $

Chord is also passes through $ (3,0) $

$ \Rightarrow \quad \frac{y-3}{x-0}=\frac{3-0}{0-3} $

$ y-3=-x \Rightarrow x+y=3 $

Put $ y=3-x $ in parabola $ x^{2}=12 y $

$ x^{2}=36-12 x $

$ x^{2}+12 x-36=0 $

$ (x-6)^{2}=0 \Rightarrow x=6,6 $

$ \Rightarrow $ Sum of reciprocal of Absicca

$ =\frac{1}{6}+\frac{1}{6}=\frac{1}{3} $

Given $ y^{2}=9 x $, normal drawn at $ P(9,9) $.

Slope of tangent $ 2 y \frac{d y}{d x}=9 $

$ \frac{d y}{d x}=\frac{9}{2 y} \Rightarrow\left(\frac{d y}{d x}\right)_{(9,9)}=\frac{1}{2} $

Slope of normal $ =-2 $

$ \left[\because m_{1} m_{2}=-1\right] $

Equation of normal

$ (y-9)=-2(x-9) \Rightarrow y=-2 x+27 $

Point $ Q(a, b) $ lies on both normal and parabola.

$ \begin{array}{l} \text { So, }(-2 x+27)^{2}=9 x \\\\ \Rightarrow 4 x^{2}-117 x+729=0 \\\\ x=\frac{117 \pm 45}{8} \Rightarrow x_{1}=20.25 \text { and } 9 \end{array} $

Now, $ x=20.25 $ they $ y=-2(20.25)+27 $

$ y=-135 $

Thus, $ Q(20.25,-135) $ i.e. $ (a, b) $

Now, $ 2 a+b \Rightarrow 2(20.25)+(13.5) $

$ \Rightarrow \quad 40.5-13.5 \Rightarrow 27 $

$ \therefore \quad 2 a+b=27 $

Given, $P$ and $Q$ are the extremities of a focal chord of the parabola

$ y^2=4 a x $

Also, $P=(9,9)$ and $Q=(p, q)$

General point on the parabola

$ y^2=4 a x \text { is }\left(a t^2, 2 a t\right) $

Now, $a t^2=9$ and $2 a t=9$

$ \Rightarrow a=\frac{9}{4} \text { and } t=2 $

$\because P$ and $Q$ are the extremities of a focal chord.

$ \begin{array}{ll} \therefore & p=a\left(-\frac{1}{t}\right)^2 \text { and } q=2 a\left(-\frac{1}{t}\right) \\\\ \Rightarrow & p=\frac{9}{4}\left(-\frac{1}{2}\right)^2 \text { and } q=2\left(\frac{9}{4}\right)\left(-\frac{1}{2}\right) \\\\ \Rightarrow & p=\frac{9}{16} \text { and } q=-\frac{9}{4} \\\\ \therefore & p-q=\frac{9}{16}-\left(-\frac{9}{4}\right)=\frac{9}{16}+\frac{9}{4}=\frac{45}{16} \end{array} $

The equation of normal to the parabola $y^2=4 a x$ is given by

$ y=m x-2 a m-a m^3 $

Here, $a=1$ and point is $(9,6)$

On putting $a=1, x=9$ and $y=6$ in the equation

$ \begin{aligned} & y=m x-2 a m-a m^3, \text { we get } \\\\ & 6=9 m-2 m-m^3 \Rightarrow m^3-7 m+6=0 \\\\ & \Rightarrow m=-3,1,2 \end{aligned} $

There are 3 distinct value of $m$.

So, the number of normals will be 3 .

We have, focus: $(2,3)$ directrix : $x-y+3=0$

Tangent drawn at the vertex of parabola is the equation of $A C$.

$\because A C$ is parallel to directrix.

$\therefore$ equation of $A C$ is $x-y+k=0 \ldots(i)$

and length of $B F=\frac{O F}{2}=\frac{1}{2}\left|\frac{2-3+3}{\sqrt{2}}\right|$

$B F=\frac{1}{\sqrt{2}}$

From Eq. (i), $B F=\left|\frac{2-3+k}{\sqrt{2}}\right|=\frac{1}{\sqrt{2}}$

$\Rightarrow k=2$

$\therefore$ Equation of tangent is $x-y+2=0$

We have, $y^2=8 x \Rightarrow a=2$

Any tangent to parabola is

$ y=m x+\frac{a}{m} \Rightarrow m x-y+\frac{2}{m}=0 $

If it is a tangent to the circle $x^2+y^2=2$, then perpendicular from centre $(0,0)$ is equal to radius $\sqrt{2}$.

$ \begin{aligned} & \therefore \frac{\left|\frac{2}{m}\right|}{\sqrt{m^2+1}}=\sqrt{2} \Rightarrow \frac{4}{m^2}=2\left(m^2+1\right) \\\\ & \Rightarrow m^4+m^2-2=0 \\\\ & \Rightarrow\left(m^2+1\right)\left(m^2-1\right)=0 \Rightarrow m= \pm 1 \end{aligned} $

Hence, the common tangents are.

$ \begin{aligned} & y= \pm(x+2) \\\\ & x+y+2=0 \text { or } x-y+2=0 \end{aligned} $

Since, $-\frac{a}{b}>0$, so in this case,

equation of tangent : $x-y+2=0$

On comparing with $a x+b y+2=0$, we get $a=1, b=-1$

So, $3 a^2+2 b+1=3-2+1=2$

$ \begin{aligned} &\text { We have, }\\\\ &\begin{aligned} & 25\left[(x-2)^2+(y+5)^2\right]=(3 x+4 y-1)^2 \\\\ & \Rightarrow \quad(x-2)^2+(y+5)^2=\left[\frac{3 x+4 y-1}{\sqrt{3^2+4^2}}\right]^2 \\\\ & \Rightarrow \quad S L=L F . \end{aligned} \end{aligned} $

So, focus is $(2,-5)$

and directrix is $3 x+4 y-1=0 \quad \ldots \text { (i) }$

Equation of $P F$ is

$ y+5=\frac{4}{3}(x-2) \quad \left[\because m_{P F}=\frac{4}{3}\right] $

$ \begin{aligned} & 3 x+4 y-1=0 \\\\ & 4 x-3 y-23=0 \quad \ldots \text { (ii) } \end{aligned} $

On solving Eqs. (i) and (ii), we get $P\left(\frac{19}{5},-\frac{13}{5}\right) \Rightarrow$ vertex $O$ is $\left(\frac{29 *}{10},-\frac{38}{10}\right)$

Now, equation of $L L^{\prime}$ :

$ \begin{aligned} 3 x+4 y+14 & =0 \end{aligned} \quad \ldots \text { (ii) } $

$\left(\because m_{L L^{\prime}}=-\frac{3}{4}\right)$

Clearly, $\left(-\frac{2}{5},-\frac{16}{5}\right)$ on parabola as well as on $L L^{\prime}$. So; one end of latus rectum is $\left(-\frac{2}{5},-\frac{16}{5}\right)$

$\therefore$ Length of latus rectum $=2\left(\sqrt{\left(-\frac{2}{5}-2\right)^2+\left(-\frac{16}{5}+5\right)^2}\right)$ $=2 \times 3=6$

$ \begin{aligned} &\\ &\begin{aligned} & \text { } y^2=16 x \\ & A(1,-4) \equiv A\left(4 t^2, 8 t\right) \Rightarrow t=-1 / 2 \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { Other coordinate will be }\\ &B\left(\frac{4}{t^2}, \frac{-8}{t}\right)=(16,16) \end{aligned} $

$ \begin{aligned} &\text { Now, equation of normal at } B(16,16)\\ &\begin{aligned} & y=-t_1 x+2 a t_1+a t_1^3 \\ \Rightarrow \quad & y=-2 x+16+32 \\ \Rightarrow \quad & 2 x+y-48=0 \end{aligned}\left[\because t_1=2, a=4\right] \end{aligned} $

$ y^2=12 x $

$ A\left(3 t^2, 6 t\right), B\left(3 t^2,-6 t\right) $

$ \begin{aligned} &O A B \text { is an equilateral triangle }\\ &\begin{aligned} & \quad O A=O B=A B \\ & \Rightarrow \quad\left(3 t^2-0\right)^2+(6 t-0)^2=(12 t)^2 \\ & \Rightarrow \quad 9 t^4=108 t^2 \\ & \Rightarrow \quad t=2 \sqrt{3} \\ & \Rightarrow \quad \text { Side of triangle }=12 t=24 \sqrt{3} \mathrm{~cm} \\ & \text { Area }=\frac{\sqrt{3}}{4}(24 \sqrt{3})^2=432 \sqrt{3} \mathrm{sq} \text { units } \end{aligned} \end{aligned} $

$\because x-2 y+k=0$

$\Rightarrow y=\frac{x}{2}+\frac{k}{2}$ is tangent to the parabola

$ y^2-4 x-4 y+8=0 $

So, slope of tangent $=y^{\prime}=\frac{1}{2}$

From the equation of parabola,

$ \begin{aligned} & 2 y\left(\frac{d y}{d x}\right)-4-4\left(\frac{d y}{d x}\right)+0=0 \\ & \Rightarrow \quad \frac{d y}{d x}=\frac{4}{2 y-4}=\frac{2}{y-2} \end{aligned} $

So, $\frac{2}{y-2}=\frac{1}{2} \Rightarrow y=6$

Put the value of $y$ in equation of parabola

$ \begin{aligned} & (6)^2-4 x-4 \times 6+8=0 \\ & \Rightarrow 4 x=20 \Rightarrow x=5 \end{aligned} $

That means, point $(5,6)$ lies on the tangent.

So, $6=\frac{5}{2}+\frac{k}{2} \Rightarrow \frac{7}{2}=\frac{k}{2}$

$ \Rightarrow \quad k=7 $

$y^2=5 x, a=\frac{5}{4} \Rightarrow$ Equation of latus rectum, $x=5 / 4$ and $x^2=5 y, a=5 / 4$

⇒ Equation of directrix, $y=-5 / 4$ Now, intersection points of both parabolas are $(0,0)$ and $(5,5)$ So, line will be $y=x$

$ \begin{aligned} & \text { So, } A=\left(\frac{5}{4}, \frac{5}{4}\right), B=\left(\frac{5}{4}, \frac{-5}{4}\right) \\ & C=\left(\frac{-5}{4}, \frac{-5}{4}\right) \\ & \because \angle A B C=90^{\circ} \end{aligned} $

Then, area $(\triangle A B C)=\frac{1}{2}(A B) \times(B C)$

$ \begin{aligned} & A B=\frac{5}{4}+\frac{5}{4}=\frac{10}{4}=5 / 2 \\ & B C=5 / 2 \end{aligned} $

Therefore, area $(\triangle A B C)=\frac{1}{2} \times \frac{5}{2} \times \frac{5}{2}$

$ =\frac{25}{8} \text { sq units } $

The equation of tangent to the parabola $y^2=8 x$ at $\left(x_0, y_0\right)$ is

$ \begin{aligned} & y_0=4\left(x+x_0\right) \text { or } 2 x-\frac{y_0}{2} y+2 x_0=0 \\ & \text { So, }-\frac{y_0}{2}=3 \\ & \Rightarrow y_0=-6 \text { and } 2 x_0=n \end{aligned} $

On putting $y=-6$ into the equation $y^2=8 x$, we get

$ (-6)^2=8 x_0 \Rightarrow x_0=\frac{9}{2} $

Thus, $\quad 2 x_0=n$

$ \begin{aligned} & \Rightarrow \quad 2\left(\frac{9}{2}\right)=n \Rightarrow n=9 \\ & \text { As, } n=9 \Rightarrow \quad(2 n, 4 \sqrt{n}) \equiv(18,12) \end{aligned} $

We know that, the equation of the normal to the parabola $y^2=4 a x$ at the point ( $x_1, y_1$ ) is given by

$ y-y_1=-\frac{y_1}{2 a}\left(x-x_1\right) $

∴ The required equation be

$ y-12=-\frac{12}{2(2)}(x-18) \text { or } 3 x+y-66=0 $

Given, the tangent $a x-y+c=0$, makes an acute angle with the positive $X$-axis in the positive direction. So, $a>0$. The line $a x-y+c=0$ is tangent to the circle $x^2+y^2=1$.

So, the distance of this line from centre $(0,0)$ of the circle $x^2+y^2=1$ will be equal to the radius 1 .

$ \begin{array}{ll} \therefore \quad 1=\frac{|c|}{\sqrt{a^2+(-1)^2}} \\ \Rightarrow \quad|c|=\sqrt{a^2+1} \\ \Rightarrow c^2=a^2+1 \\ \text { Let } a=2 \Rightarrow a^2=4 \\ \text { and } c^2=5 \end{array} $

Now, the line $2 x-y+\sqrt{5}=0$ be the common tangent line to the circle $x^2+y^2=1$ and the parabola $y^2=8 \sqrt{5} x$

$y^2=8 \sqrt{5} x$ because the equation of tangent line to the parabola

$ \begin{aligned} & y^2=8 \sqrt{5} x \text { at }\left(\frac{\sqrt{5}}{2}, 2 \sqrt{5}\right) \text { is } 2 x-y+\sqrt{5}=0 \\ & \therefore \quad a^2 c^2=4 \times 5=20 \end{aligned} $

Given parabola, $y^2=k x$

∴ Focus $\left(\frac{k}{4}, 0\right)$

$\because P\left(2, y_1\right)$ lies on the parabola.

∵ Distance between focus and $P\left(2 y_1\right)=3$

$ \sqrt{y_1^2+\left(\frac{k}{4}-2\right)^2}=3 $

By squaring both sides, we get

$ \begin{aligned} & y_1^2+\frac{k^2}{16}+4-2\left(\frac{k}{4}\right)(2)=9 \\ \Rightarrow & 2 k+\frac{k^2}{16}+4-k=9 \Rightarrow \frac{k^2}{16}+k=5 \\ \Rightarrow & k^2+16 k-80=0 \\ \Rightarrow & (k+20)(k-4)=0 \\ \therefore & k=-20,4 \\ \because & k=-24 \quad \text { (it is not possible) } \\ \therefore & k=4 \end{aligned} $

So, point $P(2, \pm 2 \sqrt{2})$

Tangent $y( \pm 2 \sqrt{2})=4\left(\frac{x+2}{2}\right)$

$ x \pm 2 \sqrt{2} y+2=0 $

Parabola $y^2=12 x$

On comparing with $y^2=4 a x$, we have

$ a=3 $

Equation of normal having slope $m$ is

$ \begin{aligned} y & =m x-2 a m-a m^3 \\ \Rightarrow \quad y & =m x-6 m-3 m^3 \end{aligned} $

Normals are passing through $P(8,0)$.

$ \begin{aligned} &\begin{aligned} & \therefore \quad 8 m-6 m-3 m^3=0 \Rightarrow 2 m-3 m^3=0 \\ & \Rightarrow \quad m\left(2-3 m^2\right)=0 \\ & \Rightarrow \quad m=0, m= \pm \frac{\sqrt{2}}{\sqrt{3}} \end{aligned}\\ &\text { Slope of non-horizontal normals }\\ &\begin{aligned} & m_1=\frac{\sqrt{2}}{\sqrt{3}} \text { and } m_2=-\frac{\sqrt{2}}{\sqrt{3}} \\ & \tan \theta=\left|\frac{m_2-m_1}{1+m_1 m_2}\right|=\frac{2 \frac{\sqrt{2}}{\sqrt{3}}}{1-\frac{2}{3}}=2 \sqrt{6} \end{aligned} \end{aligned} $

The given equation of parabola is

$ y^2=4 x $

Thus, normal equation to the parabola $y^2=4 x$ is

$ y=m x-2 m-m^3 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

(Since, any normal equation to the parabola $y^2=4 a x$ is $\left.y=m x-2 a m-a m t^2\right)$

Thus, Eq. (i), passes through $(5,0)$.

$ \begin{aligned} & \Rightarrow 0=5 m-2 m-m^3 \\ & \Rightarrow m\left(m^2+2-5\right)=0 \\ & \Rightarrow \text { Either } m=0 \text { or } m^2=3 \\ & \quad m= \pm \sqrt{3} \end{aligned} $

Thus, equation of normal passing through $(5,0)$ and whose slope is $\sqrt{3}$.

$ \begin{array}{ll} & (y-0)=\sqrt{3}(x-5) \\ \Rightarrow & \sqrt{3} x-y-5 \sqrt{3}=0 \end{array} $

Equation of parabola is $y^2=16 x\,\,\,\,\,(given)…(i) $

and the given equation of circle is

$ x^2+y^2=8 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

So, centre is $(0,0)$ and radius $=2 \sqrt{2}$ Now, equation of tangent of Eq. (i) is

$ y=m x+\frac{4}{m} $

Now, radius = length of perpendicular from the centre to tangent

$ \begin{array}{ll} \Rightarrow & 2 \sqrt{2}=\left|\frac{0+0-\frac{4}{m}}{\sqrt{1+m^2}}\right| \\ \Rightarrow & 2 \sqrt{2}=\left|\frac{-\frac{4}{m}}{\sqrt{1+m^2}}\right| \\ \Rightarrow & 8\left(1+m^2\right)=\frac{16}{m^2} \\ \Rightarrow & m^4+m^2-2=0 \\ \Rightarrow & m^4+2 m^2-m^2-2=0 \\ \Rightarrow & \left(m^2+2\right)\left(m^2-1\right)=0 \\ \therefore & m^2-1=0 \\ \therefore & m= \pm 1 \end{array} \quad\left(\because m^2+2 \neq 0\right) $

When, $m=1, y=x+4$

When, $m=-1, y=-x-4$

$\therefore$ Line of intersection

$ \begin{aligned} & x^2+y^2-6 x-6 y+13=0 \\ & x^2+y^2-8 y+9=0 \end{aligned} $

$ \begin{array}{ll} (-)(-) & (+)-(-) \\ \hline & -6 x+2 y+4=0 \\ & 3 x-y-2=0 \end{array} $

$ \begin{aligned} \therefore 9 x-3 y & =6 ; \\ x+3 y & =0 \\ 10 x & =6[\text { By adding above equation }] \\ x & =\frac{3}{5} ; y=-\frac{1}{5} \\ 0 & =\frac{a+\frac{3}{5}}{2} \Rightarrow a=-\frac{3}{5} \\ 0 & =\frac{b-\frac{1}{5}}{2} \Rightarrow b=\frac{1}{5} \\ S(a, b) & =\left(\frac{-3}{5}, \frac{1}{5}\right) \end{aligned} $

To determine the angle $\theta$ between two tangents drawn from the point $(-1, -2)$ to the parabola $y^2 = 4x$, we follow these steps:

Equation of the Tangent:

The general form of the tangent to the parabola $y^2 = 4ax$ is given by:

$ y = mx + \frac{a}{m} $

Here, $a = 1$, so the tangent equation is:

$ y = mx + \frac{1}{m} $

Substitute the Point:

Given that the point $(-1, -2)$ lies on the tangent, substitute $x = -1$ and $y = -2$ into the tangent equation:

$ -2 = -m + \frac{1}{m} $

Solve for $m$:

Rearrange and solve for $m$:

$ -2m = -m^2 + 1 \\ m^2 - 2m + 1 = 0 + 2 = 2 \\ (m - 1)^2 = 2 $

Solving $(m - 1)^2 = 2$ gives:

$ m - 1 = \pm \sqrt{2} $

Hence, the slopes of the tangents are:

$ m_1 = 1 + \sqrt{2} \quad \text{and} \quad m_2 = 1 - \sqrt{2} $

Difference of Slopes:

Calculate the difference:

$ m_1 - m_2 = (1 + \sqrt{2}) - (1 - \sqrt{2}) = 2\sqrt{2} $

Product of Slopes:

Calculate the product:

$ m_1 m_2 = (1 + \sqrt{2})(1 - \sqrt{2}) = 1^2 - (\sqrt{2})^2 = 1 - 2 = -1 $

Find $\tan \theta$:

The formula for $\tan \theta$ between two lines with slopes $m_1$ and $m_2$ is:

$ \tan \theta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right| $

Substitute the values:

$ \tan \theta = \left|\frac{2\sqrt{2}}{1 - 1}\right| = \left|\frac{2\sqrt{2}}{0}\right| $

Since the denominator is zero, $\tan \theta$ is undefined.

Conclusion:

The angle $\theta$ is $\frac{\pi}{2}$ radians, indicating that the tangents are perpendicular.

Therefore, $\theta = \frac{\pi}{2}$.

Given equation of curve,

$ \begin{aligned} & x^2-4 x+4 y-8=0 \\ \Rightarrow \quad & x^2-4 x=-4 y+8 \\ \Rightarrow \quad & x^2-4 x+4=-4 y+8+4 \\ \Rightarrow \quad & (x-2)^2=-4 y+12 \\ \Rightarrow \quad & (x-2)^2=-4(y-3) \end{aligned} $

Now, let $x-2=X$ and $y-3=Y$

$ \therefore \quad X^2=-4 Y $

On comparing Eq. (i) with $X^2=-4 a Y$, we get $4 a=4$

$ \Rightarrow \quad a=1 $

Vertex of Eq. (ii),

$ \begin{aligned} & & X & =0 \text { and } Y=0 \\ \Rightarrow & & x-2 & =0 \text { and } y-3=0 \\ \Rightarrow & & x & =2 \text { and } y=3 \end{aligned} $

∴ Vertex of Eq.(i) is $(2,3)$.

Focus of Eq. (ii) $X=0$ and $Y=-a$

$ \begin{array}{lc} \Rightarrow & x-2=0 \text { and } y-3=-1 \\ \Rightarrow & x=2 \text { and } y=2 \end{array} $

∴ Focus of Eq. (i) is $(2,2)$.

One end of the latusrectum of Eq. (ii)

$ \begin{aligned} & & X & = \pm 2 a \text { and } Y=-a \\ \Rightarrow & & x-2 & = \pm 2 \text { and } y-3=-1 \\ \Rightarrow & & x & =0,4 \text { and } y=2 \end{aligned} $

∴ One end of the latusrectum of Eq. (i) is $(4,2)$ or $(0,2)$.

Equation of directrix of Eq. (ii),

$ \begin{aligned} &\,\,\,\,\,\,\,\, Y=a & \\ \Rightarrow & y-3 =1 \Rightarrow y=4 \end{aligned} $

and equation of axis of Eq. (ii) $X=0$

$ \begin{array}{lrl} \Rightarrow & x-2 & =0 \\ \Rightarrow & x & =2 \end{array} $

∴ Point of intersection of the axis and direction is $(2,4)$.

$ y^2=\frac{8}{a} x \quad[a>0] \ldots(\mathrm{i}) $

$ y^2=4\left(\frac{2}{a}\right) x $

One end of focal chord is $(1,4)$. On putting $x=1$ and $y=4$ in Eq. (i),

$ \begin{array}{ll} & 16=\frac{8}{a} \times 1 \\ \Rightarrow & a=\frac{1}{2} \\ \therefore & y^2=\frac{8}{\left(\frac{1}{2}\right)} x \\ \Rightarrow & y^2=16 x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{array} $

Comparing $y^2=16 x$ to $y^2=4 A x$, we have

$ \begin{aligned} & & 4 A & =16 \\ \Rightarrow & & A & =4 \end{aligned} $

At $(1,4)$,

$ \begin{aligned} & 2 A t_1=y \\ & \Rightarrow \quad 2 \times 4 \times t_1=4 \Rightarrow t_1=\frac{1}{2} \\ & \text { Length of focal chord } =a\left(t+\frac{1}{t}\right)^2 \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, =4\left(\frac{1}{2}+2\right)^2 \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, =4\left(\frac{25}{4}\right)=25 \end{aligned} $

Given parabola, $y^2=8 x, a=2$

Let $P\left(x_1, y_1\right), Q\left(x_2, y_2\right)$

Equation of $A F: y-0=\frac{2-0}{1-2}(x-2)$

$ y=-2 x+4 $

$ \Rightarrow \quad 2 x+y=4 $

Taking intersection of curve and line $A F$,

$ \begin{aligned} & & (4-2 x)^2 & =8 x \\ & \Rightarrow & (2-x)^2 & =2 x \\ & \Rightarrow & x^2-4 x+4 & =2 x \\ & \Rightarrow & x^2-6 x+4 & =0 \\ & \therefore & x_1+x_2 & =6 \end{aligned} $

Given, focal distance $=3$

$ \begin{aligned} 2 t^2+a & =3 \\ \Rightarrow \quad 2 t^2+2 & =3 \\ t & = \pm \frac{1}{\sqrt{2}} \end{aligned} $

$ \begin{aligned} &\text { Given equation is }\\ &\begin{aligned} & x^2-4 y+6 x+15=0 \\ & \Rightarrow \quad x^2+6 x+9=4 y-15+9 \\ & \Rightarrow \quad(x+3)^2=4 y-6 \\ & \Rightarrow \quad(x+3)^2=4\left(y-\frac{3}{2}\right) \\ & \therefore \alpha=-3 \text { and } \beta=\frac{3}{2} \text { and } 8 a=4 \Rightarrow a=\frac{1}{2} \\ & \text { Now, } 2 \alpha+8 \beta^2=2(-3)+8 \times \frac{9}{4} \\ & \quad=-6+18=12 \end{aligned} \end{aligned} $

Given, equation of parabola

$ \begin{aligned} & y^2=16 x \\ & L L^{\prime}=4 a=16 \end{aligned} $

End points of latusrectum is $\left(a_1 \pm 2 a\right)$ Coordinates of $L$ is $(4,8)$ and $L^{\prime}$ is $(4,-8)$.

The end points of focal length is

$ \left(a t^2, 2 a t\right) \text { and }\left(\frac{a}{t^2},-\frac{2 a}{t}\right) $

Coordinate of $P$ is $(1,4)$

$ \begin{array}{ll} \Rightarrow & a t^2=1, \quad 2 a t=4 \\ \Rightarrow & 4 t^2=1, \quad 2 \times 4 \times t=4 \\ \Rightarrow & t^2=\frac{1}{4}, \quad t=\frac{1}{2} \\ \therefore & t=1 / 2 \end{array} $

∴ Coordinate of

$ Q\left(\frac{4}{\left(1 / 2^2\right.},-\frac{2 \times 4}{(1 / 2)}\right)=(16,-16) $

$ \begin{aligned} L Q & =\sqrt{(4-16)^2+(8+16)^2}=\sqrt{144+576} \\ & =\sqrt{720}=12 \sqrt{5} \end{aligned} $

Given, equation of parabola

$ y^2=32 x $

Here, $a=8$

The end point of focal chord is $\left(a t^2, 2 a t\right)$ and $\left(\frac{a}{t^2},-\frac{2 a}{t}\right)$.

$ \begin{array}{rlrl} \Rightarrow & a t^2 =\frac{1}{2} \text { and } 2 a t=4 \\ \Rightarrow & 8 \times t^2 =\frac{1}{2} \text { and } 2 \times 8 \times t=4 \\ \Rightarrow & t^2 =\frac{1}{16} \text { and } 16 t=4 \\ \Rightarrow & t= \pm \frac{1}{4} \text { and } t=\frac{1}{4} \\ & \therefore t =1 / 4 \end{array} $

Coordinate of $Q$ is $\left(\frac{8}{\left(\frac{1}{4}\right)^2}, \frac{-2 \times 8}{\frac{1}{4}}\right)$

$ =(128,-64) $

Coordinate of $S$ is $(8,0)$.

$ \begin{aligned} S Q & =\sqrt{(8-128)^2+(0+64)^2} \\ & =\sqrt{14400+4096}=\sqrt{18496}=136 \end{aligned} $

Let the point $P$ be $(h, k)$.

So, $\sqrt{(h-a)^2+(k-0)^2}=\frac{|h+k|}{\sqrt{1^2+1^2}}$

$ h^2+k^2+a^2-2 h a=\frac{h^2+k^2+2 h k}{2} $

$ 2 h^2+2 k^2+2 a^2-4 a h=h^2+k^2+2 h k $

∴ Locus of point $P$ is

$ x^2+y^2-2 x y-4 a x+2 a^2=0 $

Given, $y^2+4 y+8 x-2=0$

$ \begin{array}{r} y^2+4 y+4+8 x-6=0 \\ \left(y+2^2+2(4 x-3)=0\right. \\ y+2=0,4 x-3=0 \\ y=-2, x=3 / 4 \end{array} $

∴ Required point is $\left(\frac{3}{4},-2\right)$.

Statement I

Let $P(x, y)$ be any point on parabola. Distance from $P$ to focus = Perpendicular distance from $P$ to direction

$ \begin{aligned} & \sqrt{(x-2)^2+(y-3)^2}=\frac{|x+2 y+5|}{\sqrt{1^2+2^2}} \\ & 5\left(x^2+y^2-4 x-6 y+4+9\right)=x^2+4 y^2 +25+4 x y+20 y+10 x \\ & 4 x^2+y^2-4 x y-30 x-50 y+40=0 \end{aligned} $

∴ Statement I is true.

Statement II

$ \begin{aligned} x^2-4 x+16 y+52 & =0 \\ x^2-4 x+4 & =-16 y-48 \\ (x-4)^2 & =-16(y+3) \end{aligned} $

Comparison with $X^2=4 a Y$, we get vertex(2, - 3)

Equation of directrix is $Y=a$

$ y+3=4 \Rightarrow y=1 \text { or } y-1=0 $

∴ Statement II is wrong.

$ \begin{aligned} &\text { Given, } x=-2+2 t^2, y=2+4 t\\ &\begin{aligned} \Rightarrow & t=\frac{y-2}{4} \\ & x=-2+2\left(\frac{y-2}{4}\right)^2 \\ \Rightarrow & x=-2+\frac{y^2+4-4 y}{8} \\ & 8 x=-16+y^2+4-4 y \\ \Rightarrow & y^2-8 x-4 y-12=0 \end{aligned} \end{aligned} $

Given, equation is

$ \begin{array}{lr} & 2 x^2+5 y-6 x+1=0 \\ \Rightarrow & 2 x^2-6 x+\frac{9}{2}+5 y+1-\frac{9}{2}=0 \\ \Rightarrow & 2\left(x-\frac{3}{2}\right)^2=-5\left(y-\frac{7}{10}\right) \\ \Rightarrow & \left(x-\frac{3}{2}\right)^2=\frac{-5}{2}\left(y-\frac{7}{10}\right) \\ \Rightarrow & X^2=\frac{-5}{2} Y \end{array} $

where $X=x-\frac{3}{2}$ and $Y=y-\frac{7}{10}$

For vertex, on putting $X=0, Y=0$

$ \Rightarrow \quad x=3 / 2 \text { and } y=7 / 10 $

∴ Vertex $=(3 / 2,7 / 10)$

For focus, $X=0$ and $Y=-a$

$\Rightarrow x=\frac{3}{2}$ and $y-\frac{7}{10}=\frac{-5}{8}$

$\Rightarrow y=\frac{3}{40}$

∴ Focus $\equiv(3 / 2,3 / 40)$

As the axis of parabola is along the line $y=x$, the coordinates of focus and vertex are ( $h, h$ ) and ( $k, k$ ) respectively.

Distance of vertex from origin is $\sqrt{2}$.

$ \begin{array}{rlr} \Rightarrow & k^2+k^2 & =2 \\ \Rightarrow & k & =1 \end{array} $

Distance of focus from origin $=2 \sqrt{2}$

$ \begin{aligned} & & h^2+h^2 & =8 \Rightarrow h^2=4 \\ \Rightarrow & & h & =2 \end{aligned} $

Vertex $\equiv(1,1)$, focus $\equiv(2,2)$

Directrix passes through $(0,0)$ and perpendicular to $y=x$

∴ Directrix is $x+y=0$

∴ Parabola is

$ \begin{aligned} & (x-2)^2+(y-2)^2=\frac{(x+y)^2}{2} \\ & \Rightarrow \quad(x-y)^2+16=8(x+y) \end{aligned} $

which is satisfied by $x=(t+1)^2$,

$ y=(t-1)^2 $

Given, $y^2=16 x \Rightarrow a=4$

Let $B$ be $\left(x_1, 12\right)$ lies on parabola.

$ (12)^2=16 x_1 \Rightarrow x_1=\frac{12 \times 12}{16}=9 $

Point $B(9,12)$

Equation of focal chord passing through $(4,0)$ and ( 2,2 )

$ \begin{array}{cc} & y-0=\frac{2-0}{2-4}(x-4) \\ \Rightarrow & y=-x+4 \\ & x+y=4 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{array} $

Equation of double ordinate :

$ x=9 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Point of intersection of Eqs. (i) and (ii), $(9,-5)$.

$ \begin{aligned}Given,\,\, & y^2=16 x \quad a=4 \\ & P(4,8) \equiv\left(a t_1^2, 2 a t_1\right) \\ & 2 a t_1=8 \\ & 8 t_1=8, \Rightarrow t_1=1 \end{aligned} $

If $P Q$ is a focal chord, then

$ \begin{aligned} t_1 t_2=-1 \Rightarrow+1 \times t_2=-1 & \\ t_2=-1 & \\ Q\left(a t_2^2, 2 a t_2\right) & \equiv Q(4,-8) \\ R(16,16) & \equiv\left(a t_3^2, 2 a t_3\right) \\ 2 a t_3 & =16 \\ 8 t_3 & =16 \Rightarrow t_3=2 \end{aligned} $

If $R T$ is a focal chord, then $t_3 t_4=-1$

$ \begin{aligned} & 2 \times t_4=-1 \Rightarrow t_4=\frac{-1}{2} \\ & T\left(a t_4^2, 2 a t_4\right) \equiv T(1,-4) \\ & Q T=\sqrt{3^2+4^2}=5 \end{aligned} $

Given equation of parabola, $y^2=16 x$ and all the vertices are of an equilateral triangle with one of them coincides with the vertex of parabola

Since, given parabola is symmetrical about $X$-axis

So, $\angle A O M=30^{\circ}$

In $\triangle A O M, \tan 30^{\circ}=\frac{A M}{O M}=\frac{8 t}{4 t^2}$

$ \Rightarrow \quad \frac{1}{\sqrt{3}}=\frac{2}{t} \Rightarrow t=2 \sqrt{3} $

∴ Coordinate of point $A$ lying on parabola is

$ \left(4(2 \sqrt{3})^2, 8(2 \sqrt{3})\right) \equiv(48,16 \sqrt{3}) $

∴ Distance

$ \begin{aligned} O A & =\sqrt{\left(x_A-x_0\right)^2+\left(y_A-y_0\right)^2} \\ & =\sqrt{(48-0)^2+(16 \sqrt{3}-0)^2} \\ O A & =32 \sqrt{3} \end{aligned} $

∴ Length of the side of equilateral triangle $=32 \sqrt{3}$.